Suppose $ x+y+z=0 $. Show that $ \frac{x^5+y^5+z^5}{5}=\frac{x^2+y^2+z^2}{2}\times\frac{x^3+y^3+z^3}{3} $. [duplicate]

How to show that they are equal? All I can come up with is using symmetric polynomials to express them, or using some substitution to simplify this identity since it is symmetric and homogeneous but they are still too complicated for one to work out during the exam. So I think there should exist some better approaches to handle this identity without too much direct computation.

In addition, this identity is supposed to be true:

$$ \frac{x^7+y^7+z^7}{7}=\frac{x^2+y^2+z^2}{2}\times\frac{x^5+y^5+z^5}{5} .$$

First we can use the identities

$$ (x+y+z)^2=x^2+y^2+z^2+2(yz+zx+xy) $$

and

$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-yz-zx-xy)$$

As $x+y+z=0$, we have $x^2+y^2+z^2=-2(yz+zx+xy)$ and $x^3+y^3+z^3=3xyz$.

So, R.H.S is $-xyz(yz+zx+xy)$.

From the above results, we have

\begin{align*} (x^3+y^3+z^3)(x^2+y^2+z^2)&=x^5+y^5+z^5+x^3(y^2+z^2)+y^3(z^2+x^2)+z^3(x^2+y^2)\\ (3xyz)(-2yz-2zx-2xy)&=x^5+y^5+z^5+x^3[(y+z)^2-2yz]\\ &\quad +y^3[(z+x)^2-2zx]+z^3[(x+y)^2-2xy]\\ -6xyz(yz+zx+xy)&=x^5+y^5+z^5+x^5-2x^3yz+y^5-2xy^3z+z^5-2xyz^3\\ -6xyz(yz+zx+xy)&=2(x^5+y^5+z^5)-2xyz(x^2+y^2+z^2)\\ -6xyz(yz+zx+xy)&=2(x^5+y^5+z^5)+4xyz(yz+zx+xy)\\ x^5+y^5+z^5&=-5xyz(yz+zx+xy) \end{align*}

Hence, L.H.S. is also $-xyz(yz+zx+xy)$.

Write $S_n=x^n+y^n+z^n$. Then $S_0=3$, you are given $S_1=0$ and are charged to prove that $S_5=(5/6)S_2S_3$. Define $$F(t)=\sum_{n=0}^\infty S_nt^n.$$ Then $$ F(t)=\frac1{1-xt} + \frac1{1-yt} + \frac1{1-zt} = \frac{3+2e_1t+e_2t^2} {1-e_1t+e_2t^2-e_3t^3} $$ where $e_1=x+y+z$, $e_2=xy+xz+yz$ and $e_3=xyz$. Then $e_1=0$ so \begin{align} F(t)&=\frac{3+e_2t^2} {1+e_2t^2-e_3t^3}=(3+e_2t^2)\sum_{k=0}^\infty(-1)^k(e_2t^2-e_3t^3)^k\\ &=(3+e_2t^2)(1-e_2t^2+e_3t^3+e_2^2t^4-2e_2e_3t^5+\cdots)\\ &=3-2e_2t^2+3e_3t^3+2e_2^2t^4-5e_2e_3t^5+\cdots. \end{align} Therefore $S_2=-2e_2$, $S_3=3e_3$ and $S_5=-5e_2e_3$ etc.

By Newton's identities, with $e_k$ standing for the elementary symmetric polynomials and $p_k$ for the sum of $k^{th}$ powers, and using that $e_1=0\,$:

$$ \begin{align} p_1 &= e_1 &&= 0\\ p_2 &= e_1p_1-2e_2 &&= -2e_2 \\ p_3 &= e_1p_2 -e_2p_1 + 3e_3 &&= 3e_3 \\ p_4 &= e_1p_3 - e_2p_2+e_3p_1-4e_4 &&= 2 e_2^2 \\ p_5 &= e_1p_4-e_2p_3+e_3p_2-e_4p_1+5e_5 &&= -5e_2e_3 \end{align} $$

It follows that:

$$\color{blue}{\frac{p_5}{5}} = -e_2e_3 = \frac{-2e_2}{2} \cdot \frac{3e_3}{3} = \color{blue}{\frac{p_2}{2} \cdot \frac{p_3}{3}} $$

If $x=0$, the equality is trivial. So, it is sufficient to consider $x\ne 0$. Let $y=ax, z=abx$. Then: $$x+ax+abx=0 \Rightarrow ab=-1-a.$$ Then: $$\frac{x^5+a^5x^5+(ab)^5x^5}{5}=\frac{x^2+a^2x^2+(ab)^2x^2}{2}\cdot \frac{x^3+a^3x^3+(ab)^3x^3}{3} \Rightarrow$$ $$\frac{1+a^5+(-1-a)^5}{5}=\frac{1+a^2+(-1-a)^2}{2}\cdot \frac{1+a^3x^3+(-1-a)^3}{3} \Rightarrow$$ $$-a-2a^2-2a^3-a^4=(1+a+a^2)(-a-a^2).$$